Find the real values of $x$ that satisfy the equation $7[x]+23\{x\}=191$

Question

For any real number $x$, $[x]$ denotes the largest integer less than or equal to $x$ (i.e. floor function) and $ \{x\}=x-[x]$ .Then, the number of real solutions of the equation $$7[x]+23\{x\}=191$$ are.

My Attempt:

I used $ \{x\}=x-[x]$

This gives us,

$$23x-16[x]=191$$

as,$16[x]$ is an integer, $23x$ also must be an integer.

How to proceed next?

Answer 1

Hints:

• $23x-16[x]=191$ is a nice idea, but $7x+16\{x\}=191$ may be more useful

• Using $0 \le \{x\} \lt 1$, can you put an upper bound on $7x$? A lower bound?

• If you knew $[x]$, could you find $\{x\}$ and so $x$?

• How many possible values of $[x]$ are there? Do they all give a value of $x$ which works?

Answer 2

Adding the solution here for future references.

$$7[x]+23\{x\}=191$$

We know that, $0\leq23{\{x\}}\lt23$,

i.e. as long as $$168\lt7[x]\leq191$$

We can always find a $23{x}$ (one and only one) in $[[x],[x]+1)$ such that,

$$7[x]+23\{x\}=191$$

This gives us $[x]=25,26,27$

as $[x]=24$ gives $7[x]=168$, this will just lie on the boundary.

So, 3 solutions.

Answer 3

$$\mathrm{7}\left[{x}\right]+\mathrm{23}\left\{{x}\right\}=\mathrm{191} \\ $$ $$\left[{x}\right]=\frac{\mathrm{191}−\mathrm{23}\left\{{x}\right\}}{\mathrm{7}} \\ $$ $$\left[{x}\right]=\frac{\mathrm{191}−\mathrm{23}\left[\mathrm{0};\mathrm{1}\right)}{\mathrm{7}} \\ $$ $$\left[{x}\right]=\left[\mathrm{25};\mathrm{27}\right] \\ $$ $$\mathrm{7}\left[{x}\right]+\mathrm{23}\left({x}−\left[{x}\right]\right)=\mathrm{191} \\ $$ $${x}=\frac{\mathrm{191}+\mathrm{16}\left[{x}\right]}{\mathrm{23}} \\ $$ $${x}=\left\{\frac{\mathrm{591}}{\mathrm{23}},\frac{\mathrm{607}}{\mathrm{23}},\frac{\mathrm{623}}{\mathrm{23}}\right\} \\ $$

Answer 4

Here my attempt...

$\mathrm{7}\left[{x}\right]+\mathrm{23}\left\{{x}\right\}=\mathrm{191} \\ $ $\left[{x}\right]+\frac{\mathrm{23}\left\{{x}\right\}−\mathrm{2}}{\mathrm{7}}=\mathrm{27} \\ $ $\left\{{x}\right\}\in\left\{\frac{\mathrm{2}}{\mathrm{23}},\frac{\mathrm{9}}{\mathrm{23}},\frac{\mathrm{16}}{\mathrm{23}}\right\} \\ $ ${x}=\mathrm{27}−\frac{\mathrm{23}\left\{{x}\right\}−\mathrm{2}}{\mathrm{7}}+\left\{{x}\right\} \\ $ ${x}\in\left\{\mathrm{27}\frac{\mathrm{2}}{\mathrm{23}},\mathrm{26}\frac{\mathrm{9}}{\mathrm{23}},\mathrm{25}\frac{\mathrm{16}}{\mathrm{23}}\right\} \\ $

Hope this acceptable...